算法13------集合所有的子集或者字元串所有子串

1、題目：

給定字元串s=‘abc'，所有子集：'a','b','c','ab','ac','bc','abc'。

所有子串：’a','b','c','ab','bc','abc'

2、所有子集代碼：

（1）位運算：(輸入s為字元串）（輸出subStr為疊代器，子集清單）

def SubSet(s): 
  N = len(s) 
  for i in range(2**N): 
    subStr = [] 
    for j in range(N): 
      if(i >> j ) % 2 == 1: 
        subStr.append(s[j]) 
    yield subStr      

(2)回溯法:

def subSet(nums):
    if not nums:
        return []
    res = []
    def helper(index , tmp , m):
        res.append(tmp)
        for i in range(index , m):
            helper(i+1,tmp + [nums[i]],m)
    helper(0,[],len(nums))
    return res

if __name__=='__main__':
    nums = ['a','b','c']
    print(subSet(nums))      

（3）Python的extend功能

def subSet(nums):
    result = [[]]
    for x in nums:
        result.extend([subset + [x] for subset in result])
    return result      

3、所有子串代碼：s為字元串，ss為子串清單

ss=[s[i:i + x + 1] for x in range(len(s)) for i in range(len(s) - x)]      

轉載于:https://www.cnblogs.com/Lee-yl/p/9004115.html